Explicit homeomorphism between the plane and the plane minus a closed half-line.












1












$begingroup$


Let $r$ be a closed half-line in $mathbb{R}^2$ and $delta$ a positive real number. I want to find an explicit homeomorphism $phi:mathbb{R}^2tomathbb{R}^2-r$ such that $phi(X)=X$ for all $X$ such that $d(X,r)geqdelta.$



I can suppose the half-line is ${(x,y)inmathbb{R}^2:xgeq0,y=0}.$ Moreover, if we put $$B={Xinmathbb{R}^2:d(X,r)<delta},$$ we must have $phi(B)=B-r.$



I don’t know how to write this homeomorphism. I thought I can deform this initial configuration into something other, but I don’t have any other ideas.










share|cite|improve this question











$endgroup$












  • $begingroup$
    ${(x,y)inmathbb{R}^2:xgeq0}$ is the closed right half plane. Do you mean ${(x,y)inmathbb{R}^2:xgeq0 , y =0}$?
    $endgroup$
    – Paul Frost
    Dec 27 '18 at 22:48










  • $begingroup$
    Yes, I have corrected the mistake.
    $endgroup$
    – MatP
    Dec 28 '18 at 4:56
















1












$begingroup$


Let $r$ be a closed half-line in $mathbb{R}^2$ and $delta$ a positive real number. I want to find an explicit homeomorphism $phi:mathbb{R}^2tomathbb{R}^2-r$ such that $phi(X)=X$ for all $X$ such that $d(X,r)geqdelta.$



I can suppose the half-line is ${(x,y)inmathbb{R}^2:xgeq0,y=0}.$ Moreover, if we put $$B={Xinmathbb{R}^2:d(X,r)<delta},$$ we must have $phi(B)=B-r.$



I don’t know how to write this homeomorphism. I thought I can deform this initial configuration into something other, but I don’t have any other ideas.










share|cite|improve this question











$endgroup$












  • $begingroup$
    ${(x,y)inmathbb{R}^2:xgeq0}$ is the closed right half plane. Do you mean ${(x,y)inmathbb{R}^2:xgeq0 , y =0}$?
    $endgroup$
    – Paul Frost
    Dec 27 '18 at 22:48










  • $begingroup$
    Yes, I have corrected the mistake.
    $endgroup$
    – MatP
    Dec 28 '18 at 4:56














1












1








1





$begingroup$


Let $r$ be a closed half-line in $mathbb{R}^2$ and $delta$ a positive real number. I want to find an explicit homeomorphism $phi:mathbb{R}^2tomathbb{R}^2-r$ such that $phi(X)=X$ for all $X$ such that $d(X,r)geqdelta.$



I can suppose the half-line is ${(x,y)inmathbb{R}^2:xgeq0,y=0}.$ Moreover, if we put $$B={Xinmathbb{R}^2:d(X,r)<delta},$$ we must have $phi(B)=B-r.$



I don’t know how to write this homeomorphism. I thought I can deform this initial configuration into something other, but I don’t have any other ideas.










share|cite|improve this question











$endgroup$




Let $r$ be a closed half-line in $mathbb{R}^2$ and $delta$ a positive real number. I want to find an explicit homeomorphism $phi:mathbb{R}^2tomathbb{R}^2-r$ such that $phi(X)=X$ for all $X$ such that $d(X,r)geqdelta.$



I can suppose the half-line is ${(x,y)inmathbb{R}^2:xgeq0,y=0}.$ Moreover, if we put $$B={Xinmathbb{R}^2:d(X,r)<delta},$$ we must have $phi(B)=B-r.$



I don’t know how to write this homeomorphism. I thought I can deform this initial configuration into something other, but I don’t have any other ideas.







general-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 4:56







MatP

















asked Dec 27 '18 at 18:00









MatPMatP

1367




1367












  • $begingroup$
    ${(x,y)inmathbb{R}^2:xgeq0}$ is the closed right half plane. Do you mean ${(x,y)inmathbb{R}^2:xgeq0 , y =0}$?
    $endgroup$
    – Paul Frost
    Dec 27 '18 at 22:48










  • $begingroup$
    Yes, I have corrected the mistake.
    $endgroup$
    – MatP
    Dec 28 '18 at 4:56


















  • $begingroup$
    ${(x,y)inmathbb{R}^2:xgeq0}$ is the closed right half plane. Do you mean ${(x,y)inmathbb{R}^2:xgeq0 , y =0}$?
    $endgroup$
    – Paul Frost
    Dec 27 '18 at 22:48










  • $begingroup$
    Yes, I have corrected the mistake.
    $endgroup$
    – MatP
    Dec 28 '18 at 4:56
















$begingroup$
${(x,y)inmathbb{R}^2:xgeq0}$ is the closed right half plane. Do you mean ${(x,y)inmathbb{R}^2:xgeq0 , y =0}$?
$endgroup$
– Paul Frost
Dec 27 '18 at 22:48




$begingroup$
${(x,y)inmathbb{R}^2:xgeq0}$ is the closed right half plane. Do you mean ${(x,y)inmathbb{R}^2:xgeq0 , y =0}$?
$endgroup$
– Paul Frost
Dec 27 '18 at 22:48












$begingroup$
Yes, I have corrected the mistake.
$endgroup$
– MatP
Dec 28 '18 at 4:56




$begingroup$
Yes, I have corrected the mistake.
$endgroup$
– MatP
Dec 28 '18 at 4:56










2 Answers
2






active

oldest

votes


















0












$begingroup$

Let us proceed in various steps.



(1) It clearly suffices to show that for any $a > 0$ there exists a homeomorphism $h_a : left( [-a,infty) times [-a,a] right) setminus r to [-a,infty) times [-a,a]$ which is the identity on the boundary $B_a$ of the strip $S_a = [-a,infty) times [-a,a]$.



(2) It suffices to consider $a = 1$. Define $g_a : [-1,infty) times [-1,1] to [-a,infty) times [-a,a], g_a(x,y) = (ax,ay)$. This is a homeomorphism (its inverse is $g_{1/a}$) such that $g_a(r) = r$ and $g_a(B_1) = B_a$. If we have found $h_1$, then take $h_a = g_a circ h_1 circ g_{1/a}$.



(3) Define a map $p: [-1,1] times [-1,1] to [-1,1] times [-1,1]$ as follows. For each $y in [-1,1]$ let $p_y : [-1,1] to [-1,1]$ denote the piecewise linear map such that $p_y(-1) = -1$, $p_y(-lvert y rvert) = 1 - 2 lvert y rvert$, $p_y(1) = 1$. Then define $p(x,y) = (p_y(x),y)$. Explicitly
$$p(x,y) =
begin{cases}
1 +2 x & -1 le x le - lvert y rvert \
1 - 2 lvert y rvert + dfrac{2 lvert y rvert}{1 + lvert y rvert}(x + lvert y rvert) & - lvert y rvert le x le 1
end{cases}
$$

It is an easy exercise to verify that $p$ is continuous. Moreover, $p$ is the identity on the boundary $D$ of the square $Q = [-1,1] times [-1,1]$. It maps $[0,1] times { 0 }$ to the point $(0,1)$ and restricts to a continuous bijection $p ': Q' setminus r' to Q'$, where $Q' = [-1,1) times [-1,1]$ and $r' = [0,1) times { 0 }$. Since $Q$ is compact, $p$ is a closed map and therefore a quotient map. Let $U subset Q' setminus r'$ be open. Since $Q' setminus r'$ is open in $Q$, $U$ is open in $Q$. We have $p^{-1}(p(U)) = U$, hence $p(U)$ is open in $Q$. Thus $p'(U) = p(U) = p(U) cap Q'$ is open in $Q'$ which shows that $p'$ is an open map, i.e. a homeomorphism.



(4) Let us finally construct the homeomorphism $h_1$ as in (2). Define a homeomorphism $psi : [-1,1) to [-1,infty)$ by $psi(x) = x$ for $x le 0$ and $psi(x) = dfrac{x}{1-x}$ for $x ge 0$. Then $psi' = psi times id : Q' to S_1$ is a homeomorphism such that $psi'(r') = r$. It maps the boundary of $Q'$ onto the boundary of $S_1$ and restricts to a homeomorphism $phi: Q' setminus r' to S_1 setminus r$. Now define
$$h_1 = phi circ p' circ phi^{-1} .$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @MatP Any reaction would be nice. If you are not satisfied with an answer, you should tell it to author, and you can also downvote. On the other hand, if the answer is helpful, you should accept it. See math.stackexchange.com/help/someone-answers. No reaction also leaves the question in the uanswered queue which is confusing for the users. If you click at "Unanswered" in the upper left corne of this page, you will read something like "235,857 questions with no upvoted or accepted answers ". If you look at these questions, you will see that a great many are actually answered.
    $endgroup$
    – Paul Frost
    Jan 7 at 14:20



















0












$begingroup$

Using polar coordinates (r,t) where

{ (r,0) : 0 <= r } is the removed closed ray,

map (r,t) to (r,$pi$ + t/2).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In polar coordinates $(r,t)$ stands for $re^{it}$ with $t in (0,2pi)$. But then $pi +t/2 in (pi,2pi)$, i.e. the image of your map is the open lower half plane.
    $endgroup$
    – Paul Frost
    Dec 28 '18 at 9:27










  • $begingroup$
    You got it. Now finish the homeomorphism.
    $endgroup$
    – William Elliot
    Dec 28 '18 at 22:06










  • $begingroup$
    Yes, but it will be not a homeomorphism as required in the question.
    $endgroup$
    – Paul Frost
    Dec 28 '18 at 22:48











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054219%2fexplicit-homeomorphism-between-the-plane-and-the-plane-minus-a-closed-half-line%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Let us proceed in various steps.



(1) It clearly suffices to show that for any $a > 0$ there exists a homeomorphism $h_a : left( [-a,infty) times [-a,a] right) setminus r to [-a,infty) times [-a,a]$ which is the identity on the boundary $B_a$ of the strip $S_a = [-a,infty) times [-a,a]$.



(2) It suffices to consider $a = 1$. Define $g_a : [-1,infty) times [-1,1] to [-a,infty) times [-a,a], g_a(x,y) = (ax,ay)$. This is a homeomorphism (its inverse is $g_{1/a}$) such that $g_a(r) = r$ and $g_a(B_1) = B_a$. If we have found $h_1$, then take $h_a = g_a circ h_1 circ g_{1/a}$.



(3) Define a map $p: [-1,1] times [-1,1] to [-1,1] times [-1,1]$ as follows. For each $y in [-1,1]$ let $p_y : [-1,1] to [-1,1]$ denote the piecewise linear map such that $p_y(-1) = -1$, $p_y(-lvert y rvert) = 1 - 2 lvert y rvert$, $p_y(1) = 1$. Then define $p(x,y) = (p_y(x),y)$. Explicitly
$$p(x,y) =
begin{cases}
1 +2 x & -1 le x le - lvert y rvert \
1 - 2 lvert y rvert + dfrac{2 lvert y rvert}{1 + lvert y rvert}(x + lvert y rvert) & - lvert y rvert le x le 1
end{cases}
$$

It is an easy exercise to verify that $p$ is continuous. Moreover, $p$ is the identity on the boundary $D$ of the square $Q = [-1,1] times [-1,1]$. It maps $[0,1] times { 0 }$ to the point $(0,1)$ and restricts to a continuous bijection $p ': Q' setminus r' to Q'$, where $Q' = [-1,1) times [-1,1]$ and $r' = [0,1) times { 0 }$. Since $Q$ is compact, $p$ is a closed map and therefore a quotient map. Let $U subset Q' setminus r'$ be open. Since $Q' setminus r'$ is open in $Q$, $U$ is open in $Q$. We have $p^{-1}(p(U)) = U$, hence $p(U)$ is open in $Q$. Thus $p'(U) = p(U) = p(U) cap Q'$ is open in $Q'$ which shows that $p'$ is an open map, i.e. a homeomorphism.



(4) Let us finally construct the homeomorphism $h_1$ as in (2). Define a homeomorphism $psi : [-1,1) to [-1,infty)$ by $psi(x) = x$ for $x le 0$ and $psi(x) = dfrac{x}{1-x}$ for $x ge 0$. Then $psi' = psi times id : Q' to S_1$ is a homeomorphism such that $psi'(r') = r$. It maps the boundary of $Q'$ onto the boundary of $S_1$ and restricts to a homeomorphism $phi: Q' setminus r' to S_1 setminus r$. Now define
$$h_1 = phi circ p' circ phi^{-1} .$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @MatP Any reaction would be nice. If you are not satisfied with an answer, you should tell it to author, and you can also downvote. On the other hand, if the answer is helpful, you should accept it. See math.stackexchange.com/help/someone-answers. No reaction also leaves the question in the uanswered queue which is confusing for the users. If you click at "Unanswered" in the upper left corne of this page, you will read something like "235,857 questions with no upvoted or accepted answers ". If you look at these questions, you will see that a great many are actually answered.
    $endgroup$
    – Paul Frost
    Jan 7 at 14:20
















0












$begingroup$

Let us proceed in various steps.



(1) It clearly suffices to show that for any $a > 0$ there exists a homeomorphism $h_a : left( [-a,infty) times [-a,a] right) setminus r to [-a,infty) times [-a,a]$ which is the identity on the boundary $B_a$ of the strip $S_a = [-a,infty) times [-a,a]$.



(2) It suffices to consider $a = 1$. Define $g_a : [-1,infty) times [-1,1] to [-a,infty) times [-a,a], g_a(x,y) = (ax,ay)$. This is a homeomorphism (its inverse is $g_{1/a}$) such that $g_a(r) = r$ and $g_a(B_1) = B_a$. If we have found $h_1$, then take $h_a = g_a circ h_1 circ g_{1/a}$.



(3) Define a map $p: [-1,1] times [-1,1] to [-1,1] times [-1,1]$ as follows. For each $y in [-1,1]$ let $p_y : [-1,1] to [-1,1]$ denote the piecewise linear map such that $p_y(-1) = -1$, $p_y(-lvert y rvert) = 1 - 2 lvert y rvert$, $p_y(1) = 1$. Then define $p(x,y) = (p_y(x),y)$. Explicitly
$$p(x,y) =
begin{cases}
1 +2 x & -1 le x le - lvert y rvert \
1 - 2 lvert y rvert + dfrac{2 lvert y rvert}{1 + lvert y rvert}(x + lvert y rvert) & - lvert y rvert le x le 1
end{cases}
$$

It is an easy exercise to verify that $p$ is continuous. Moreover, $p$ is the identity on the boundary $D$ of the square $Q = [-1,1] times [-1,1]$. It maps $[0,1] times { 0 }$ to the point $(0,1)$ and restricts to a continuous bijection $p ': Q' setminus r' to Q'$, where $Q' = [-1,1) times [-1,1]$ and $r' = [0,1) times { 0 }$. Since $Q$ is compact, $p$ is a closed map and therefore a quotient map. Let $U subset Q' setminus r'$ be open. Since $Q' setminus r'$ is open in $Q$, $U$ is open in $Q$. We have $p^{-1}(p(U)) = U$, hence $p(U)$ is open in $Q$. Thus $p'(U) = p(U) = p(U) cap Q'$ is open in $Q'$ which shows that $p'$ is an open map, i.e. a homeomorphism.



(4) Let us finally construct the homeomorphism $h_1$ as in (2). Define a homeomorphism $psi : [-1,1) to [-1,infty)$ by $psi(x) = x$ for $x le 0$ and $psi(x) = dfrac{x}{1-x}$ for $x ge 0$. Then $psi' = psi times id : Q' to S_1$ is a homeomorphism such that $psi'(r') = r$. It maps the boundary of $Q'$ onto the boundary of $S_1$ and restricts to a homeomorphism $phi: Q' setminus r' to S_1 setminus r$. Now define
$$h_1 = phi circ p' circ phi^{-1} .$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @MatP Any reaction would be nice. If you are not satisfied with an answer, you should tell it to author, and you can also downvote. On the other hand, if the answer is helpful, you should accept it. See math.stackexchange.com/help/someone-answers. No reaction also leaves the question in the uanswered queue which is confusing for the users. If you click at "Unanswered" in the upper left corne of this page, you will read something like "235,857 questions with no upvoted or accepted answers ". If you look at these questions, you will see that a great many are actually answered.
    $endgroup$
    – Paul Frost
    Jan 7 at 14:20














0












0








0





$begingroup$

Let us proceed in various steps.



(1) It clearly suffices to show that for any $a > 0$ there exists a homeomorphism $h_a : left( [-a,infty) times [-a,a] right) setminus r to [-a,infty) times [-a,a]$ which is the identity on the boundary $B_a$ of the strip $S_a = [-a,infty) times [-a,a]$.



(2) It suffices to consider $a = 1$. Define $g_a : [-1,infty) times [-1,1] to [-a,infty) times [-a,a], g_a(x,y) = (ax,ay)$. This is a homeomorphism (its inverse is $g_{1/a}$) such that $g_a(r) = r$ and $g_a(B_1) = B_a$. If we have found $h_1$, then take $h_a = g_a circ h_1 circ g_{1/a}$.



(3) Define a map $p: [-1,1] times [-1,1] to [-1,1] times [-1,1]$ as follows. For each $y in [-1,1]$ let $p_y : [-1,1] to [-1,1]$ denote the piecewise linear map such that $p_y(-1) = -1$, $p_y(-lvert y rvert) = 1 - 2 lvert y rvert$, $p_y(1) = 1$. Then define $p(x,y) = (p_y(x),y)$. Explicitly
$$p(x,y) =
begin{cases}
1 +2 x & -1 le x le - lvert y rvert \
1 - 2 lvert y rvert + dfrac{2 lvert y rvert}{1 + lvert y rvert}(x + lvert y rvert) & - lvert y rvert le x le 1
end{cases}
$$

It is an easy exercise to verify that $p$ is continuous. Moreover, $p$ is the identity on the boundary $D$ of the square $Q = [-1,1] times [-1,1]$. It maps $[0,1] times { 0 }$ to the point $(0,1)$ and restricts to a continuous bijection $p ': Q' setminus r' to Q'$, where $Q' = [-1,1) times [-1,1]$ and $r' = [0,1) times { 0 }$. Since $Q$ is compact, $p$ is a closed map and therefore a quotient map. Let $U subset Q' setminus r'$ be open. Since $Q' setminus r'$ is open in $Q$, $U$ is open in $Q$. We have $p^{-1}(p(U)) = U$, hence $p(U)$ is open in $Q$. Thus $p'(U) = p(U) = p(U) cap Q'$ is open in $Q'$ which shows that $p'$ is an open map, i.e. a homeomorphism.



(4) Let us finally construct the homeomorphism $h_1$ as in (2). Define a homeomorphism $psi : [-1,1) to [-1,infty)$ by $psi(x) = x$ for $x le 0$ and $psi(x) = dfrac{x}{1-x}$ for $x ge 0$. Then $psi' = psi times id : Q' to S_1$ is a homeomorphism such that $psi'(r') = r$. It maps the boundary of $Q'$ onto the boundary of $S_1$ and restricts to a homeomorphism $phi: Q' setminus r' to S_1 setminus r$. Now define
$$h_1 = phi circ p' circ phi^{-1} .$$






share|cite|improve this answer











$endgroup$



Let us proceed in various steps.



(1) It clearly suffices to show that for any $a > 0$ there exists a homeomorphism $h_a : left( [-a,infty) times [-a,a] right) setminus r to [-a,infty) times [-a,a]$ which is the identity on the boundary $B_a$ of the strip $S_a = [-a,infty) times [-a,a]$.



(2) It suffices to consider $a = 1$. Define $g_a : [-1,infty) times [-1,1] to [-a,infty) times [-a,a], g_a(x,y) = (ax,ay)$. This is a homeomorphism (its inverse is $g_{1/a}$) such that $g_a(r) = r$ and $g_a(B_1) = B_a$. If we have found $h_1$, then take $h_a = g_a circ h_1 circ g_{1/a}$.



(3) Define a map $p: [-1,1] times [-1,1] to [-1,1] times [-1,1]$ as follows. For each $y in [-1,1]$ let $p_y : [-1,1] to [-1,1]$ denote the piecewise linear map such that $p_y(-1) = -1$, $p_y(-lvert y rvert) = 1 - 2 lvert y rvert$, $p_y(1) = 1$. Then define $p(x,y) = (p_y(x),y)$. Explicitly
$$p(x,y) =
begin{cases}
1 +2 x & -1 le x le - lvert y rvert \
1 - 2 lvert y rvert + dfrac{2 lvert y rvert}{1 + lvert y rvert}(x + lvert y rvert) & - lvert y rvert le x le 1
end{cases}
$$

It is an easy exercise to verify that $p$ is continuous. Moreover, $p$ is the identity on the boundary $D$ of the square $Q = [-1,1] times [-1,1]$. It maps $[0,1] times { 0 }$ to the point $(0,1)$ and restricts to a continuous bijection $p ': Q' setminus r' to Q'$, where $Q' = [-1,1) times [-1,1]$ and $r' = [0,1) times { 0 }$. Since $Q$ is compact, $p$ is a closed map and therefore a quotient map. Let $U subset Q' setminus r'$ be open. Since $Q' setminus r'$ is open in $Q$, $U$ is open in $Q$. We have $p^{-1}(p(U)) = U$, hence $p(U)$ is open in $Q$. Thus $p'(U) = p(U) = p(U) cap Q'$ is open in $Q'$ which shows that $p'$ is an open map, i.e. a homeomorphism.



(4) Let us finally construct the homeomorphism $h_1$ as in (2). Define a homeomorphism $psi : [-1,1) to [-1,infty)$ by $psi(x) = x$ for $x le 0$ and $psi(x) = dfrac{x}{1-x}$ for $x ge 0$. Then $psi' = psi times id : Q' to S_1$ is a homeomorphism such that $psi'(r') = r$. It maps the boundary of $Q'$ onto the boundary of $S_1$ and restricts to a homeomorphism $phi: Q' setminus r' to S_1 setminus r$. Now define
$$h_1 = phi circ p' circ phi^{-1} .$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 28 '18 at 11:34

























answered Dec 28 '18 at 11:28









Paul FrostPaul Frost

11.6k3935




11.6k3935












  • $begingroup$
    @MatP Any reaction would be nice. If you are not satisfied with an answer, you should tell it to author, and you can also downvote. On the other hand, if the answer is helpful, you should accept it. See math.stackexchange.com/help/someone-answers. No reaction also leaves the question in the uanswered queue which is confusing for the users. If you click at "Unanswered" in the upper left corne of this page, you will read something like "235,857 questions with no upvoted or accepted answers ". If you look at these questions, you will see that a great many are actually answered.
    $endgroup$
    – Paul Frost
    Jan 7 at 14:20


















  • $begingroup$
    @MatP Any reaction would be nice. If you are not satisfied with an answer, you should tell it to author, and you can also downvote. On the other hand, if the answer is helpful, you should accept it. See math.stackexchange.com/help/someone-answers. No reaction also leaves the question in the uanswered queue which is confusing for the users. If you click at "Unanswered" in the upper left corne of this page, you will read something like "235,857 questions with no upvoted or accepted answers ". If you look at these questions, you will see that a great many are actually answered.
    $endgroup$
    – Paul Frost
    Jan 7 at 14:20
















$begingroup$
@MatP Any reaction would be nice. If you are not satisfied with an answer, you should tell it to author, and you can also downvote. On the other hand, if the answer is helpful, you should accept it. See math.stackexchange.com/help/someone-answers. No reaction also leaves the question in the uanswered queue which is confusing for the users. If you click at "Unanswered" in the upper left corne of this page, you will read something like "235,857 questions with no upvoted or accepted answers ". If you look at these questions, you will see that a great many are actually answered.
$endgroup$
– Paul Frost
Jan 7 at 14:20




$begingroup$
@MatP Any reaction would be nice. If you are not satisfied with an answer, you should tell it to author, and you can also downvote. On the other hand, if the answer is helpful, you should accept it. See math.stackexchange.com/help/someone-answers. No reaction also leaves the question in the uanswered queue which is confusing for the users. If you click at "Unanswered" in the upper left corne of this page, you will read something like "235,857 questions with no upvoted or accepted answers ". If you look at these questions, you will see that a great many are actually answered.
$endgroup$
– Paul Frost
Jan 7 at 14:20











0












$begingroup$

Using polar coordinates (r,t) where

{ (r,0) : 0 <= r } is the removed closed ray,

map (r,t) to (r,$pi$ + t/2).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In polar coordinates $(r,t)$ stands for $re^{it}$ with $t in (0,2pi)$. But then $pi +t/2 in (pi,2pi)$, i.e. the image of your map is the open lower half plane.
    $endgroup$
    – Paul Frost
    Dec 28 '18 at 9:27










  • $begingroup$
    You got it. Now finish the homeomorphism.
    $endgroup$
    – William Elliot
    Dec 28 '18 at 22:06










  • $begingroup$
    Yes, but it will be not a homeomorphism as required in the question.
    $endgroup$
    – Paul Frost
    Dec 28 '18 at 22:48
















0












$begingroup$

Using polar coordinates (r,t) where

{ (r,0) : 0 <= r } is the removed closed ray,

map (r,t) to (r,$pi$ + t/2).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In polar coordinates $(r,t)$ stands for $re^{it}$ with $t in (0,2pi)$. But then $pi +t/2 in (pi,2pi)$, i.e. the image of your map is the open lower half plane.
    $endgroup$
    – Paul Frost
    Dec 28 '18 at 9:27










  • $begingroup$
    You got it. Now finish the homeomorphism.
    $endgroup$
    – William Elliot
    Dec 28 '18 at 22:06










  • $begingroup$
    Yes, but it will be not a homeomorphism as required in the question.
    $endgroup$
    – Paul Frost
    Dec 28 '18 at 22:48














0












0








0





$begingroup$

Using polar coordinates (r,t) where

{ (r,0) : 0 <= r } is the removed closed ray,

map (r,t) to (r,$pi$ + t/2).






share|cite|improve this answer









$endgroup$



Using polar coordinates (r,t) where

{ (r,0) : 0 <= r } is the removed closed ray,

map (r,t) to (r,$pi$ + t/2).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 28 '18 at 7:24









William ElliotWilliam Elliot

8,7222820




8,7222820












  • $begingroup$
    In polar coordinates $(r,t)$ stands for $re^{it}$ with $t in (0,2pi)$. But then $pi +t/2 in (pi,2pi)$, i.e. the image of your map is the open lower half plane.
    $endgroup$
    – Paul Frost
    Dec 28 '18 at 9:27










  • $begingroup$
    You got it. Now finish the homeomorphism.
    $endgroup$
    – William Elliot
    Dec 28 '18 at 22:06










  • $begingroup$
    Yes, but it will be not a homeomorphism as required in the question.
    $endgroup$
    – Paul Frost
    Dec 28 '18 at 22:48


















  • $begingroup$
    In polar coordinates $(r,t)$ stands for $re^{it}$ with $t in (0,2pi)$. But then $pi +t/2 in (pi,2pi)$, i.e. the image of your map is the open lower half plane.
    $endgroup$
    – Paul Frost
    Dec 28 '18 at 9:27










  • $begingroup$
    You got it. Now finish the homeomorphism.
    $endgroup$
    – William Elliot
    Dec 28 '18 at 22:06










  • $begingroup$
    Yes, but it will be not a homeomorphism as required in the question.
    $endgroup$
    – Paul Frost
    Dec 28 '18 at 22:48
















$begingroup$
In polar coordinates $(r,t)$ stands for $re^{it}$ with $t in (0,2pi)$. But then $pi +t/2 in (pi,2pi)$, i.e. the image of your map is the open lower half plane.
$endgroup$
– Paul Frost
Dec 28 '18 at 9:27




$begingroup$
In polar coordinates $(r,t)$ stands for $re^{it}$ with $t in (0,2pi)$. But then $pi +t/2 in (pi,2pi)$, i.e. the image of your map is the open lower half plane.
$endgroup$
– Paul Frost
Dec 28 '18 at 9:27












$begingroup$
You got it. Now finish the homeomorphism.
$endgroup$
– William Elliot
Dec 28 '18 at 22:06




$begingroup$
You got it. Now finish the homeomorphism.
$endgroup$
– William Elliot
Dec 28 '18 at 22:06












$begingroup$
Yes, but it will be not a homeomorphism as required in the question.
$endgroup$
– Paul Frost
Dec 28 '18 at 22:48




$begingroup$
Yes, but it will be not a homeomorphism as required in the question.
$endgroup$
– Paul Frost
Dec 28 '18 at 22:48


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054219%2fexplicit-homeomorphism-between-the-plane-and-the-plane-minus-a-closed-half-line%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Aardman Animations

Are they similar matrix

“minimization” problem in Euclidean space related to orthonormal basis